This isn't a very specific question ... by four digits, do you mean four numbers? Furthermore, are the digits or numbers you use allowed to repeat?
Let's assume you meant numbers, not digits. In this case, for example, you could be trying to create a locker combination, and the numbers you can use are 1 through 20, and the combination has four "blank" spaces.
1. If the numbers can repeat, then each spot has 20 possibilities, so the total number of combinations is 20 for the first spot, 20 for the second, 20 for the third, and 20 for the fourth. That is, 20*20*20*20 = 160,000 combinations.
2. If the numbers cannot repeat, then the first spot can be taken up by any of 20 numbers, the second spot only by 19 (because you can't repeat the first number, whatever it was), the third spot by 18, and the fourth spot by any one of the 17 remaining numbers. So that makes 20*19*18*17 possibilities = 116,280 combinations.
If this is not what you meant, and you wanted digits, remember that every number above and including 10 has two digits, so you'll want to factor that into your calculations.
There are 5*4*3*2 = 120 such numbers.
120
There are 6!/3! = 120 different 3-digit numbers that can be made from these 6 digits.
10 x 9 x 8 = 720 different arrangements If the sequence doesn't matter, then there are (720/6) = 120 different bunches
120 different combinations (5 x 4 x 3 x 2).
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
There are 5*4*3*2 = 120 such numbers.
To calculate the number of 3-digit combinations using numbers 1-6, including 1 and 6, we need to consider the total number of options for each digit. Since we are including 1 and 6, there are 6 options for the hundreds place (1-6), 6 options for the tens place (1-6), and 6 options for the ones place (1-6). Therefore, the total number of 3-digit combinations is 6 x 6 x 6 = 216.
5 digit combinations for numbers 1 to 50= 50 C 5N!= 1x2x3x4x...x (N-1) x N= (50!)/(45!x5!) =(50x49x...x2x1)/((45x44x...x2x1)(5x4x3x2x1))=(50x49x48x47x46)/(5x4x3x2)=254251200/120 =21187602118760 combinations
120
Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.
There are: 10C3 = 120
120 5x4x3x2x1=120
If you can repeat the numbers, it will be 6*6*6=216. If you can't, it will be 6*5*4=120.
120
There are 6!/3! = 120 different 3-digit numbers that can be made from these 6 digits.